With permission of the author, we will demonstrate how to translate the book Euclidean Plane Geometry, written by Milan Mitrović from Slovenian into English, without modifying any of the LaTeX commands.
To achieve this, we will first split the book into chunks, each roughly a page long, then translate each chunk into English, and finally stitch them back together.
import openai
from transformers import GPT2Tokenizer
# OpenAI GPT-2 tokenizer is the same as GPT-3 tokenizer
# we use it to count the number of tokens in the text
tokenizer = GPT2Tokenizer.from_pretrained("gpt2")
with open("data/geometry_slovenian.tex", "r") as f:
text = f.read()1485565
chunks = text.split('\n\n')
ntokens = []
for chunk in chunks:
ntokens.append(len(tokenizer.encode(chunk)))
max(ntokens)Token indices sequence length is longer than the specified maximum sequence length for this model (1327 > 1024). Running this sequence through the model will result in indexing errors
1473
It turns out that a double newline is a good separator in this case, in order not to break the flow of the text. Also no individual chunk is larger than 1500 tokens. The model we will use is text-davinci-002, which has a limit of 4096 tokens, so we don't need to worry about breaking the chunks down further.
We will group the shorter chunks into chunks of around 1000 tokens, to increase the coherence of the text, and decrease the frequency of breaks within the text.
def group_chunks(chunks, ntokens, max_len=1000, hard_max_len=3000):
"""
Group very short chunks, to form approximately page long chunks.
"""
batches = []
cur_batch = ""
cur_tokens = 0
# iterate over chunks, and group the short ones together
for chunk, ntoken in zip(chunks, ntokens):
# discard chunks that exceed hard max length
if ntoken > hard_max_len:
print(f"Warning: Chunk discarded for being too long ({ntoken} tokens > {hard_max_len} token limit). Preview: '{chunk[:50]}...'")
continue
# if room in current batch, add new chunk
if cur_tokens + 1 + ntoken <= max_len:
cur_batch += "\n\n" + chunk
cur_tokens += 1 + ntoken # adds 1 token for the two newlines
# otherwise, record the batch and start a new one
else:
batches.append(cur_batch)
cur_batch = chunk
cur_tokens = ntoken
if cur_batch: # add the last batch if it's not empty
batches.append(cur_batch)
return batches
chunks = group_chunks(chunks, ntokens)
len(chunks)869
Notice that adding a sample untranslated and translated first command, where only the content of the chapter name needs to be translated, helps to get more consistent results.
The format of the prompt sent to the model consists of:
The expected output is the translated chunk of text.
def translate_chunk(chunk, engine='text-davinci-002',
dest_language='English',
sample_translation=("\poglavje{Osnove Geometrije} \label{osn9Geom}", "\poglavje{The basics of Geometry} \label{osn9Geom}")
):
prompt = f'''Translate only the text from the following LaTeX document into {dest_language}. Leave all LaTeX commands unchanged
"""
{sample_translation[0]}
{chunk}"""
{sample_translation[1]}
'''
response = openai.Completion.create(
prompt=prompt,
engine=engine,
temperature=0,
top_p=1,
max_tokens=1500,
)
result = response['choices'][0]['text'].strip()
result = result.replace('"""', '') # remove the double quotes, as we used them to surround the text
return result
print(translate_chunk(chunks[800], engine='text-davinci-002', dest_language='English'))Let $\mathcal{I}=\mathcal{S}_{AB} \circ\mathcal{S}_{CA}
\circ\mathcal{S}_{BC}$. By \ref{izoZrcdrsprq} is
$\mathcal{I}$ a mirror reflection. Let $A_1$, $B_1$ and $C_1$ be in order the center points of the lines $BC$, $AC$ and $AB$ of the triangle $ABC$.
Because it is a right triangle is $\mathcal{I}(A_1C_1)=A_1C_1$, which
means that the line $A_1C_1$ is of this mirror reflection. It is not
difficult to prove that for the point $A'_1=\mathcal{I}(A_1)$ (both
lie on the axis $A_1C_1$) is
$\overrightarrow{A_1A'_1}=3\overrightarrow{A_1C_1}$, so
$\mathcal{I}=\mathcal{G}_{3\overrightarrow{A_1C_1}}$.
\item \res{Given are the points $A$ and $B$ on the same side of the line
$p$.
Draw the line $XY$, which lies on the line $p$ and is consistent
with the given line $l$, so that the sum
$|AX|+|XY|+|YB|$ is minimal.}
Let $A'=\mathcal{G}_{\overrightarrow{MN}}(A)$ (where $M,N\in
p$ and $MN\cong l$). The point $Y$ is obtained as the intersection of the lines $p$
and $X'Y$ (see also example \ref{HeronProbl}).
\item \res{Let $ABC$ be an isosceles right triangle with a right angle at the vertex $A$. What does the composite
$\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$ represent?}
Let $p$ and $q$ be the simetrali of the sides $CA$ and $AB$ of the triangle
$ABC$. By \ref{izoZrcDrsKompSrOsn} is:
$$\mathcal{G}_{\overrightarrow{AB}}\circ
\mathcal{G}_{\overrightarrow{CA}}=
\mathcal{S}_q\circ\mathcal{S}_A\circ\mathcal{S}_A\circ\mathcal{S}_p=
\mathcal{S}_q\circ\mathcal{S}_p.$$ Because $ABC$ is an isosceles
right triangle with a right angle at the vertex $A$, the lines $p$ and $q$ are perpendicular and intersect at the center $S$
of the hypotenuse $BC$. Therefore
$\mathcal{G}_{\overrightarrow{AB}}\circ
\mathcal{G}_{\overrightarrow{CA}}=\mathcal{S}_q
\circ\mathcal{S}_p=\mathcal{S}_S$.
\item \res{In the same plane are given the lines
$a$, $b$ and $c$.
Draw the points $A\in a$ and $B\in b$
so that $\mathcal{S}_c(A)=B$.}
We can see here that this one chunk in particular translates only the text, but leaves LaTeX commands intact.
Let's now translate all the chunks in the book - this will take 2-3 hours, as we're processing requests sequentially.
dest_language = "English"
translated_chunks = []
for i, chunk in enumerate(chunks):
print(str(i+1) + " / " + str(len(chunks)))
# translate each chunk
translated_chunks.append(translate_chunk(chunk, engine='text-davinci-002', dest_language=dest_language))
# join the chunks together
result = '\n\n'.join(translated_chunks)
# save the final result
with open(f"data/geometry_{dest_language}.tex", "w") as f:
f.write(result)0 / 869 1 / 869 2 / 869 3 / 869 4 / 869 5 / 869 6 / 869 7 / 869 8 / 869 9 / 869 10 / 869 11 / 869 12 / 869 13 / 869 14 / 869 15 / 869 16 / 869 17 / 869 18 / 869 19 / 869 20 / 869 21 / 869 22 / 869 23 / 869 24 / 869 25 / 869 26 / 869 27 / 869 28 / 869 29 / 869 30 / 869 31 / 869 32 / 869 33 / 869 34 / 869 35 / 869 36 / 869 37 / 869 38 / 869 39 / 869 40 / 869 41 / 869 42 / 869 43 / 869 44 / 869 45 / 869 46 / 869 47 / 869 48 / 869 49 / 869 50 / 869 51 / 869 52 / 869 53 / 869 54 / 869 55 / 869 56 / 869 57 / 869 58 / 869 59 / 869 60 / 869 61 / 869 62 / 869 63 / 869 64 / 869 65 / 869 66 / 869 67 / 869 68 / 869 69 / 869 70 / 869 71 / 869 72 / 869 73 / 869 74 / 869 75 / 869 76 / 869 77 / 869 78 / 869 79 / 869 80 / 869 81 / 869 82 / 869 83 / 869 84 / 869 85 / 869 86 / 869 87 / 869 88 / 869 89 / 869 90 / 869 91 / 869 92 / 869 93 / 869 94 / 869 95 / 869 96 / 869 97 / 869 98 / 869 99 / 869 100 / 869 101 / 869 102 / 869 103 / 869 104 / 869 105 / 869 106 / 869 107 / 869 108 / 869 109 / 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